A291393 p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - S)(1 - 2 S).

3, 10, 29, 83, 232, 643, 1771, 4862, 13321, 36455, 99696, 272535, 744839, 2035358, 5561381, 15195075, 41515496, 113425323, 309888403, 846638398, 2313071313, 6319448079, 17265085152, 47169141487, 128868574671, 352075628734, 961888724621, 2627929220939

Offset

0,1

Comments

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A291382 for a guide to related sequences.

Links

Clark Kimberling, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (3, 1, -4, -2)

Formula

G.f.: -(((1 + x) (-3 + 2 x + 2 x^2))/((-1 + x + x^2) (-1 + 2 x + 2 x^2))).

a(n) = 3*a(n-1) + a(n-2) - 4*a(n-3) - 2*a(n-4) for n >= 5.

Mathematica

z = 60; s = x + x^2; p = (1 - s)(1 - 2s);
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291393 *)

Crossrefs

Cf. A019590, A291382.

Sequence in context: A269144 A096140 A307262 * A244615 A307062 A052976

Adjacent sequences: A291390 A291391 A291392 * A291394 A291395 A291396

Keyword

nonn,easy

Author

Clark Kimberling, Sep 06 2017

Status

approved